Oil tankers the mean number of oil tankers at a port city is eight per day. find the probability that the number of oil tankers on any given day is (a) exactly eight, (b) at most three, and (c) more than eight.

a) P(8) = 0.1395

b) P(at most three) = 0.0423684

c) P(X > 8) = 0.41

Step-by-step explanation:

Data provided in the question:

Mean, μ = 8

Now,

Probability that the number of oil tankers on any given day is

a) exactly eight

using Poisson distribution

we have

P(x) =

for x = 8

P(8) =

or

P(8) =

or

P(8) = 0.1395

b)  at most three

i.e P(0) + P(1) + P(2) + P(3)

thus,

P(0) = = 0.0003354

P(1) = = 0.002683

P(2) = = 0.01073

P(3) = = 0.02862

⇒ P(at most three) = 0.0003354 + 0.002683 + 0.01073 + 0.02862

= 0.0423684

c) more than eight.

P(X > 8) = 1 - P(X ≤ 8)

Now,

P(0) = = 0.0003354

P(1) = = 0.002683

P(2) = = 0.01073

P(3) = = 0.02862

P(4) = = 0.05725

P(5) = = 0.091603

P(6) = = 0.1221

P(7) = = 0.13958

P(8) = = 0.13758

Thus,

P(X > 8) = 1 - [ 0.0003354 + 0.002683 + 0.01073 + 0.02862 + 0.05725 + 0.091603 + 0.1221 + 0.13958 + 0.13758 ]

P(X > 8) = 1 - 0.5904814

or

P(X > 8) = 0.41

can someone answer this question plz

step-by-step explanation:

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).